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Q108P

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Fundamentals Of Physics
Found in: Page 1115

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Short Answer

White light (consisting of wavelengths from 400 nm to 700 nm ) is normally incident on a grating. Show that, no matter what the value of the grating spacing d , the second order and third order overlap.

It is proved no matter what the value of the grating spacing d, the second order and third order overlap.

See the step by step solution

Step by Step Solution

Step 1: Given data                                                                             

White light has wavelengths between

λ1=400 nm

λ2=700 nm

Step 2: Definition and concept of diffraction from a grating

An optical element with a periodic structure that divides light into a number of beams that move in diverse directions is known as a diffraction grating.

The angular distance of the mth order diffraction maxima is given by

dsinθ=mλ …(i)

Here, λis the wavelength of the incident light and d is the separation between two lines in the grating.

Step 3: Determining the angular distances of second and third order maxima

From equation (i) the angular distance of second order maxima from λ2 is

dsinθ2=2x70mmmsinθ2=1400mmd

From equation (i) the angular distance of third order maxima from λ1 is

dsinθ3=3x400mmsinθ3=1200mmd

Thus

sinθ2>sinθ3

Hence the angular distance of a part of the second order from white light is greater than the angular distance of a part of the third order. Thus the second and third order overlap.

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